[Numpy-discussion] ignore NAN in numpy.true_divide()

[questions anon]

Thanks for responding. I have tried several ways of adding the command, one
of which is:

        for i in TSFC:
                if N.any(N.isnan(TSFC)):
                        break
                else:
                        pass
but nothing is happening, is there some particular way I need to add this
command? I have posted all below:

netCDF_list=[]

for dir in glob.glob(MainFolder + '*/01/')+ glob.glob(MainFolder +
'*/02/')+ glob.glob(MainFolder + '*/12/'):
        #print dir
        for ncfile in glob.glob(dir + '*.nc'):
            netCDF_list.append(ncfile)

slice_counter=0
print netCDF_list
for filename in netCDF_list:
        ncfile=netCDF4.Dataset(filename)
        TSFC=ncfile.variables['T_SFC'][:]
        fillvalue=ncfile.variables['T_SFC']._FillValue
        TSFC=MA.masked_values(TSFC, fillvalue)
        for a in TSFC:
                if N.any(N.isnan(TSFC)):
                        break
                else:
                        pass

        for i in xrange(0,len(TSFC)-1,1):
                        slice_counter +=1
                #print slice_counter
                        try:
                                running_sum=N.add(running_sum, TSFC[i])
                        except NameError:
                                print "Initiating the running total of my
variable..."
                                running_sum=N.array(TSFC[i])

TSFC_avg=N.true_divide(running_sum, slice_counter)
N.set_printoptions(threshold='nan')
print "the TSFC_avg is:", TSFC_avg




On Tue, Dec 6, 2011 at 9:45 AM, David Cournapeau <cournape at gmail.com> wrote:

> On Mon, Dec 5, 2011 at 5:29 PM, questions anon <questions.anon at gmail.com>
> wrote:
> > Maybe I am asking the wrong question or could go about this another way.
> > I have thousands of numpy arrays to flick through, could I just identify
> > which arrays have NAN's and for now ignore the entire array. is there a
> > simple way to do this?
>
> Doing np.any(np.isnan(a)) for an array a should answer this exact question
>
> David
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